Finding common ratio and more?

**Jukixel** · January 3rd 2009, 03:28 PM

I actually don't even know what topic this is. O.o"
It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
a) Calculate the value of the common ratio r.

b) If the fourth term is 16/9, calculate the sum of the infinite series.

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.

**Danny** · January 3rd 2009, 03:45 PM

Originally Posted by Jukixel

I actually don't even know what topic this is. O.o"
It's a three part problem and I will probably ask lots of questions because I don't think I've learned this yet.

An infinite geometric series whose terms are positive has a finite sum. The ratio of the sum of the first four terms to the sum of the first two terms is 13:9.
a) Calculate the value of the common ratio r.

b) If the fourth term is 16/9, calculate the sum of the infinite series.

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.

Hints:

Let the series be $S = a + a r + a r^2 +a r^3 + a r^4 + \cdots$ . You said the series has a finite answer so it converges and thus $-1 < r < 1$

For (a) consider $\frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}$

which factors

$\frac{a(1+r)(1+r^2)}{a(1+r)} = \frac{13}{9}$ from which you can find $r$ .

Once you know $r$ you should be able to find $a$ (that's using the fact in (b)). Do you know the formula for the sum of a converging infinite geometric series?

**Jukixel** · January 3rd 2009, 03:48 PM

No, don't think I know it.

**Danny** · January 3rd 2009, 03:50 PM

Originally Posted by Jukixel

No, don't think I know it.

If $-1 < r < 1$ then the sum of

$S = a + ar + ar^2 + ar^3 + \cdots$

is

$S_{\infty} = \frac{a}{1-r}$

**skeeter** · January 3rd 2009, 03:58 PM

The ratio of the sum of the first four terms to the sum of the first two terms is 13:9
a) Calculate the value of the common ratio r.

let r be the common ratio ...

$\frac{a + ar + ar^2 + ar^3}{a + ar} = \frac{13}{9}$

$\frac{1 + r + r^2 + r^3}{1 + r} = \frac{13}{9}$

$9 + 9r + 9r^2 + 9r^3 = 13 + 13r$

$9r^3 + 9r^2 - 4r - 4 = 0$

$9r^2(r + 1) - 4(r + 1) = 0$

$(r+1)(9r^2 - 4) = 0$

$(r+1)(3r+2)(3r-2) = 0$

$r = -1$ or $r = -\frac{2}{3}$ or $r = \frac{2}{3}$

An infinite geometric series whose terms are positive has a finite sum.

$r = \frac{2}{3}$ is the only valid solution

b) If the fourth term is 16/9, calculate the sum of the infinite series.

$ar^3 = \frac{16}{9}$

sum of a finite infinite geometric series is $S = \frac{a}{1-r}$

c) Find the least value of n for which the sum to n terms differs from the sum of the infinite series by less than 0.01.

sum of the first n terms of a geometric series is $S_n = \frac{a(1-r^n)}{1-r}$

**Jukixel** · January 4th 2009, 04:23 PM

Thanks!

Thread: Finding common ratio and more?

LinkBack

Thread Tools

Display

Finding common ratio and more?

Similar Math Help Forum Discussions

find the common ratio?

common ratio

common ratio of the seqence

Common ratio problems

Common ratio problems

Search Tags